I guess this is the solution. Would be grateful if someone could verify the arguments.
The proof of the continuous embedding $W^{1,2}(0,T;\mathbb{R}^d)\hookrightarrow \mathbb{L}^2(0,T;\mathbb{R}^d)$ follows from Evans 5.9.2.
So let's begin with the assumption that we have a bounded sequence $(x_n)$ in $X:=W^{1,2}(0,T;\mathbb{R}^d)$, $\sup\limits_{n}\|x_n\|_X<c$ (i.e. we have uniform bounds).
(This assumption may be incorrect (is it??). But it actually holds in the original problem.)
For proving the compact embedding we make use of Azrela-Ascoli theorem which requires
(1) Uniform bounds on $(x_n)$ in the space $X$ which we have.
(2) Equicontinuity:
Let $0\leq s\leq t\leq T$.
\begin{equation}
\|x_n(t)-x_n(s)\|_{\mathbb{R}^d}\leq\int\limits_{s}^t\|\dot{x}_n(\sigma)\|_{\mathbb{R}^d}d\sigma\leq \|\dot{x}_n\|_{\mathbb{L}^2(0,T;\mathbb{R}^d)}(t-s)^{\frac{1}{2}}\leq C(t-s)^{\frac{1}{2}},
\end{equation}
The first inequality is due to fundamental theorem of calculus (and existence of weak derivative).
By Azrela-Ascoli we have managed to show the compactness of $(x_n)$ in the Holder space $C^{\frac{1}{2}}(0,T;\mathbb{R}^d)$. By similar arguments we can show that Holder space $C^\alpha$ is compactly embedded in $C^\beta$ where $\beta>\alpha$. Hence we are done.
Note: There is a subtlety involved in the usage of Azrela-Ascoli theorem. We require that the space $\overline{\{x_n(t): t\in[0,T]\}}$ is compact in $\mathbb{R}^d$. Therefore this embedding would have not been possible for any range space $X$ instead of $\mathbb{R}^d$.
Suppose you want to find a number $r$ whose square $r^{2}$ is $2$. That has no meaning for numerical analysis because all numbers on a computer are rational, and $\sqrt{2}$ is not rational. It wasn't until the late 1800's that Mathematicians found a logically consistent way to define a real number. But once such a beast could be defined, then one can prove that various algorithms will get you closer and closer to $r$ to $\sqrt{2}$, knowing that it has something to converge to. The existence of such a thing in the extended "real" number system became important to the discussion.
Sobolev spaces are to the ordinary differentiable functions what the real numbers are to the rational numbers. In the late 1800's it was discovered that Calculus of variations didn't have minimizing or maximizing functions. It was the same type of problem: a larger class of functions had to be considered, and the corresponding definitions of integrals had to be extended in order to make sense of and to find a unique minimizer or maximizer that would solve variational problems. So new functions spaces emerged, Lebesgue integration extended the integral expressions to new function classes, and solutions could be found. Once minimizing or maximizing functions could be found, their properties could be deduced, and it validated various algorithms used to find solutions that couldn't converge before because there was nothing to converge to.
Best Answer
No, not on $\mathbb R^n$. Take a smooth function $\phi$ with support contained in the unit ball. Consider the sequence $f_n(x) = \phi(x-3ne_1)$ where $e_1$ is a basis vector. This is a sequence of bumps going into infinity. Translation by $3ne_1$ does not change either $L^2$ norm or $W^{-1,2}$ norm (I write $W^{-1,2}$ for $(W_0^{1,2})^*$ here). Since the supports are separated, for $n\ne m$ $$\|f_n-f_m\|_{W^{-1,2}}\ge \frac{\langle f_n-f_m, f_n\rangle}{\|f_n\|_{W^{1,2}}} = \frac{\langle f_n , f_n\rangle}{\|f_n\|_{W^{1,2}}} = c $$ where $c>0$ does not depend on $n$, due to translation invariance. Thus, $(f_n)$ has no convergent subsequence in $W^{-1,2}$.
Remarks
The argument applies to other Sobolev-type spaces with translation-invariant norms. Pretty much nothing is compactly embedded on $\mathbb R^n$, unless you use a weight decaying at infinity.
On domains where the Rellich-Kondrachov theorem applies, you get compact embedding by dualizing the embedding of $W_0^{1,2}$ into $L^2$. The adjoint of a compact operator is compact.